Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

Dividing the previous equation by x:

We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

Calculate the number of moles of CO2 and water considering the same:


The total number of moles at the reactor output would be:

So, the oxygen mole fraction would be:
%
Answer:
It does not matter where the sample of water came from or how it was prepared. Its composition, like that of every other compound, is fixed.
multiply by 100.4.36×10-5cm

1.305 × (23+35.5) = 76.34 grams
Following reactions show the major product,